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Question

Two identical metal plates are given positive charge Q1 and Q2 (<Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is

A
(Q1+Q2)/2C
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B
(Q1+Q2)/C
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C
(Q1Q2)/C
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D
(Q1Q2)/(2C)
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Solution

The correct option is D (Q1Q2)/(2C)

The potential difference between the two identical metal plates is given as

C=ε0Ad

Let the surface charge density is given as

σ1=σ21=QA

The net electric field is

Enet=σ1σ22ε0

We know the potential difference is given as

V=E.d

By substituting the above values we get

V=Q1Q22C


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